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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Migratory Birds
  5. Discussions

Migratory Birds

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3640 Discussions

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  • hgreen
     + 0 comments

    Here is how I solved it using C++. Feel free to leave feedback on my solution:

    `cpp

    int migratoryBirds(vector arr) {

    // Dump each bird type in a hashmap with its frequency
std::unordered_map<int, int> map;
for(int a : arr) {
    ++map[a];
}

// Then, check each type (start from type 5)
// Return type with highest frequency (starting from end will result in highestCt having lowest ID)
int resultType = 0;
int highestCt = -1;
for(int i = 5; i > 0; --i) {
    int count = map[i];
    if(count >= highestCt) {
        highestCt = count;
        resultType = i;
    }
}

return resultType;

    }`

  • aayuuhere
     + 0 comments

    Java Solution

    using HashMaps

     public static int migratoryBirds(List<Integer> arr) {
    // Write your code here
        int curMax = 0 ;
        HashMap<Integer , Integer> map = new HashMap<>();
        for(int num : arr){
            if(map.containsKey(num)){
                int curFreq = map.get(num);
                map.put(num , curFreq+1);
                curMax = Math.max(curMax , map.get(num));
            }
            else{
                map.put(num ,1);
                curMax = Math.max(curMax,1);
            }
        }
        int res = Integer.MAX_VALUE;
        for(int key : map.keySet()){
            if(map.get(key) == curMax){
                res = Math.min(res,key);
            }
        }
        return res;
    }

  • mahendar_karnati
     + 0 comments
    public static int migratoryBirds(List<Integer> arr) {

Map<Integer,Integer> m=new HashMap<>();

    for(int i: arr){
        if(m.containsKey(i)){
            m.put(i, m.get(i)+1);
        }
        else{
            m.put(i, 1);
        }
    }
    int key=0;
    int value=0;
    for(int i:m.keySet()){
        if(m.get(i)>value){
            value=m.get(i);
            key=i;
        }
    }
    return key;

}

    }

  • sebtabbasquresh1
     + 0 comments

    Clean problem — reminds me of how I categorized menu items by popularity on the Popeyes menu site. Here's my Python take using Counter:

    from collections import Counter

def migratoryBirds(arr):
    counts = Counter(arr)
    return min([k for k, v in counts.items() if v == max(counts.values())])

  • bhupankarg2
     + 0 comments
    int migratoryBirds(vector<int> arr)
{
    int ret = 0;
    unordered_map<int, int> hMap;
    for(int i = 0; i < arr.size(); i++)
    {
        hMap[arr[i]]++;
    }
    int key1 = -1, maxOcc1 = -1;
    int key2 = -1, maxOcc2 = -1;
    for(auto it = hMap.begin(); it != hMap.end(); it++)
    {
        if(it->second > maxOcc1)
        {
            key1 = it->first;
            maxOcc1 = it->second;
            maxOcc2 = -1;
        }
        else if(it->second == maxOcc1)
        {
            key2 = it->first;
            maxOcc2 = it->second;
        }
    }
    if(maxOcc1 == maxOcc2)
    {
        if(key1 < key2) ret = key1;
        else ret = key2;
    }
    else
    {
        ret = key1;
    }
    return ret;
}